Proof.

Suppose ker⁡f⊇ker⁡ϕ1.
If f=0, then the result is trivial.
Otherwise, there exists y∈X such that f⁢(y)≠0.
By hypothesis, we also have ϕ1⁢(y)≠0.
Every z∈X can be decomposed into z=x+t⁢y
where x∈ker⁡ϕ1⊆ker⁡f, and t is a scalar.
Indeed, just set t=ϕ1⁢(z)/ϕ1⁢(y), and x=z-t⁢y.
Then we propose that

Now suppose we have ker⁡f⊇⋂i=1nker⁡ϕi
for n>1.
Restrict each of the functionals
to the subspaceW=ker⁡ϕn, so that
ker⁡f|W⊇⋂i=1n-1ker⁡ϕi|W.
By the induction hypothesis, there exist scalars λ1,…,λn-1
such that f|W=∑i=1n-1λi⁢ϕi|W.
Then ker⁡(f-∑i=1n-1λi⁢ϕi)⊇W=ker⁡ϕn, and the argument for the case n=1
can be applied anew, to obtain the final λn.
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